A reductive algebraic group with no Borel subgroup

Question

A Borel subgroup of a linear algebraic group $G$ is a subgroup $B$ of $G$ such that $G/B$ is proper and such that $B$ is minimal with respect to this property. For example, the group $B$ of upper triangular invertible matrices is a Borel subgroup of the group $GL_n$; the quotient $GL_n/B$ is the space of maximal flags in an $n$-dimensional vector space.

Over an algebraically closed field, every reductive group contains a Borel subgroup. I have read that over an arbitrary field, there are examples of reductive groups that do not contain any Borel subgroups. Can someone give an example of such a reductive group?

Answer 1

Let $k$ be a field of characteristic $p$ admitting a nontrivial central division algebra $D$. I claim that the algebraic group $G=GL_1(D)$ has no Borel subgroups. It is a general fact that the geometric unipotent radical of a Borel subgroup descends to a split unipotent group defined over $k$. (Here split unipotent means admitting a composition series with composition factors the additive group $\mathbb G_a$.) Therefore, if $G$ has a Borel subgroup then $G$ contains a copy of $\mathbb G_a$. Since $\mathbb G_a(k)$ has elements of order $p$, it would then follow that $G(k) = D^\times$ has elements of order $p$.

On the other hand, every nonzero element $\alpha$ of the group $D$ lies in a finite extension $k(\alpha)$ of $k$. In characteristic $p$, no field element can have order $p$, since the polynomial $x^p-1$ factors as $(x-1)^p$. Hence $GL_1(D)$ has no Borel subgroups.

(This is example is from Brian Conrad's notes on algebraic groups.)

Answer 2

Just to add on to user134824's nice answer, if $k$ is any field and $D$ is a $k$-central division algebra over $k$ then $D^\times$ has no parabolic subgroups (i.e. subgroups $P$ such that $D^\times/P$ is proper (equiv. projective)). I will explain the proof below when $k$ is characteristic $0$. This is almost certainly not necessary, but since I don't use the following fact for groups over characteristic $p$ often, I only know for sure/know the proof in characteristic $0$, so I'll only state it there.

The point is that the following nice fact holds:

Theorem: Let $k$ be a characteristic $0$ field (just to avoid annoying worries about difficulties of pseudo-reductivity) and let $G$ be a connected reductive group over $k$. Then, the following are equivalent:

• $G$ has no parabolic subgroups.
• $G/Z(G)$ contains no copy of $\mathbb{G}_{m,k}$.
• $G(k)$ contains no non-trivial unipotent elements.

So now, let $G=D^\times$. Suppose that $u\in G(k)=D^\times$ is unipotent. Note that $G$ embeds into $\mathrm{GL}_k(D)$, let's call this embeding $j$, via left multiplication so we have that $j(u)$ in $\mathrm{GL}_k(D)$ is unipotent. So, $(j(u)-I)^m=0$ for some $m\geqslant 1$. But, note that the embedding of groups $G\hookrightarrow \mathrm{GL}_k(D)$ arises from an embedding of algebras $j:D\hookrightarrow \mathrm{End}_k(D)$. So, we see that

$$0=(j(u)-I)^m=j\left((u-1)^m\right)$$

and since $j$ is an embedding this implies that $(u-1)^m=0$ which, since $D$ is a division algebra, implies that $u=1$. The conclusion follows.